# 7049398 ```Adapted by Peter Au, George Brown College
McGraw-Hill Ryerson
Copyright &copy; 2011 McGraw-Hill Ryerson Limited.
8.1
8.2
8.3
8.4
8.5
8.6
Null and Alternative Hypotheses and Errors in
Testing
Type I and Type II Errors and Their Probabilities
Z Tests about a Population Mean (s Known):
One-Sided Alternatives
Z Tests about a Population Mean (s Known):
Two-Sided Alternatives
t Tests about a Population Mean (s Unknown)
Z Tests about a Population Proportion
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-2
L01
• The null hypothesis, denoted H0, is a statement of
the basic proposition being tested. The statement
generally represents the status quo and is not
rejected unless there is convincing sample
evidence that it is false
• The alternative or research hypothesis, denoted
Ha, is an alternative (to the null hypothesis)
statement that will be accepted only if there is
convincing sample evidence that it is true
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-3
L01
• With a new billing system, the mean bill paying
time m is hoped
• to be less than 19.5 days
• The alternative hypothesis Ha is that the new billing system has a
mean payment time that is less than 19.5 days
• With the old billing system, the mean bill paying time m was close
to but not less than 39 days, so much more than 19.5 days
• The null hypothesis H0 is that the new billing
system has a mean payment time close to but not
less than 39 days (which is more than 19.5 days)
• One-Sided, “Less Than” Alternative
• H0: m  19.5 vs.Ha: m &lt; 19.5
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-4
L01
• An important quality characteristic is the case
hardness depth, in millimeters (mm), of the
• The optimal target value is 4.5 mm
•
Take m = 4.5 mm
• Test whether the case hardness is 4.5 mm on average (the
null) or not (the alternative)
•
Two-Sided, “Not Equal To” Alternative
•
H0: m = 4.5 vs. Ha: m  4.5
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-5
L01
• One-Sided, “Greater Than” Alternative
H0: m  m0 vs. Ha: m &gt; m0
• One-Sided, “Less Than” Alternative
H0 : m  m0 vs. Ha : m &lt; m0
• Two-Sided, “Not Equal To” Alternative
H0 : m = m0 vs. Ha : m  m0
where m0 is a given constant value (with the
appropriate units) that is a comparative value
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-6
L01
• As a result of testing H0 vs. Ha, will have to decide
either of the following decisions for the null
hypothesis H0:
• Do not reject H0
• A weaker statement than “accepting H0”
• But you are rejecting the alternative Ha
OR
• Reject H0
• A weaker statement than “accepting Ha”
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-7
L01
• In order to “test” H0 vs. Ha, use the “test statistic”
x  m0 x  m0
z

sx
s n
where m0 is the given value or the hypothesized mean
(often the claimed to be true) and x-bar is the mean of a
sample
• If the population is normal, then the test statistic z
follows a normal distribution but the parent population
DOES NOT HAVE TO BE NORMAL
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-8
L01
• In the payment time case, with m0 = 19.5, use the test
statistic:
x  19.5 x  19.5
z
sx

s
n
• If z ≤ 0, then x ≤ m0 and thus, there is evidence to support
rejecting H0 in favor of Ha
• The point estimate x indicates that m is probably less
than 19.5
• If z &gt; 0, then x &gt; m0 and thus, there is no evidence to
support rejecting H0
• The point estimate x indicates that m is probably greater
than 19.5
• The larger the value of z (the farther x is above m), the stronger the
evidence to support H0
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-9
L03
• Type I Error: Rejecting H0 when it is true
• Type II Error: Failing to reject H0 when it is false
State of Nature
Conclusion
H0 True
H0 False
Reject H0
Type I Error
Correct Decision
Do not Reject H0
Correct Decision
Type II Error
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-10
L03
• Type I Error: Rejecting H0 when it is true
• α is the probability of making a Type I error
• 1 – a is the probability of not making a Type I error
• Type II Error: Failing to reject H0 when it is false
• β is the probability of making a Type II error
• 1 – b is the probability of not making a Type II error
State of Nature
Conclusion
Reject H0
Do not Reject H0
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
H0 True
H0 False
α
1-α
1-β
β
8-11
L03
• Usually set a to a low value
• So that there is only a small chance of rejecting a true H0
• Typically, a = 0.05
• For a = 0.05, strong evidence is required to reject H0
• Usually choose a between 0.01 and 0.05
• a = 0.01 requires very strong evidence is to reject H0
• Sometimes choose a as high as 0.10
• Tradeoff between a and b
• For fixed sample size, the lower we set a, the higher is b
• The higher a, the lower b
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-12
L04
• Test hypotheses about a population mean using
the normal distribution
• These tests are called Z tests
• They require that the true value of the population
standard deviation s is known
• In most real-world situations, s is not known
• But often is estimated from s of a single sample
• When s is unknown, test hypotheses about a
population mean using the t distribution (later)
• Here, assume that we know s
• Test using a “rejection point rule”
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-13
L04
The steps are as follows:
1. State the null and alternative hypotheses
2. Specify the significance level a
3. Select the test statistic
4. Determine the rejection rule for deciding whether
or not to reject H0
5. Collect the sample data and calculate the value of
the test statistic
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
7. Interpret the statistical results in managerial terms
and assess their practical importance
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-14
L04
• Cable TV companies claim that their full cable packages
cost on average \$50 per month
• A marketing company wants to demonstrate that it is
significantly higher
1. State the null and alternative hypotheses
• H0: m  50
Ha: m &gt; 50
• μ is the hypothesized mean
• We modify the hypothesis to
H0: m = 50
Ha: m &gt; 50
• since if there is evidence to reject H0 at μ = 50 then there is
evidence to reject it at μ ≤ 50
2. Specify the significance level a
• a = 0.05
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-15
L04
3. Select the test statistic
• Use the test statistic:
z
x  50 x  50

sx
s n
• A positive value of this this test statistic results
from a sample mean that is greater than \$50
•
Which provides evidence against H0 in favor of Ha
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-16
L04
4. Determine the rejection rule for deciding whether or
not to reject H0
• To decide how large the test statistic must be to reject
H0 by setting the probability of a Type I error to a, do
the following:
• The probability a is the area in the right-hand tail of the
standard normal curve (one tailed test)
• Use the normal table to find the point za (called the
rejection or critical point)
• za is the point under the standard normal curve that gives
a right-hand tail area equal to a
• Since a = 0.05 in the cable case, the rejection point is
za = z0.05 = 1.645
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-17
L04
4. Continued
• Reject H0 in favor of Ha if the test statistic z is
greater than the rejection point za
• This is the rejection rule
• In the cable subscriptions example, the rejection
rule is to reject H0 if the calculated test statistic
z is &gt; 1.645
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-18
L04
5. Collect the sample data and calculate the value
of the test statistic
• Assume that s is known and s = \$1.65
• For a sample of n = 40, we observed a value of
x  \$50.575
x  50 50.575  50
z

 2.20
s n
1.65 40
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-19
L05
L06
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
• Compare the value of the test statistic to the
rejection point according to the rejection rule
• In the cable case,
z = 2.20 (test statistic)
z0.05 = 1.645 (rejection point)
2.20 &gt; 1.645
• Therefore reject H0: m ≤ 50 in favor of Ha: m &gt; 50 at
the 0.05 significance level
• Have rejected H0 by using a test that allows only a 5% chance
of wrongly rejecting H0
• This result is “statistically significant” at the 0.05 level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-20
L05
L06
7. Interpret the statistical results in managerial
terms and assess their practical importance
• We conclude that it appears as though the mean
full cable subscription package is more than \$50
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-21
L07
• Testing H0: m = 50 versus Ha: m &gt; 50 for (a) α = 0.05
and (b) α = 0.01
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8-22
L07
• At a = 0.01, the rejection point is z0.01 = 2.33
• In the previous example, the test statistic
z = 2.20 is &lt; z0.01 = 2.33
• Therefore, cannot reject H0 in favor of Ha at the
a = 0.01 significance level
• This is the opposite conclusion reached with
a = 0.05
• So, the smaller we set a, the larger is the rejection
point, and the stronger is the statistical evidence
that is required to reject the null hypothesis H0
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-23
L08
• The p-value or the observed level of significance is the probability of
the obtaining the sample result if the null hypothesis H0 is true
• The p-value is a conditional probability
• It measures the level of support there is for H0 given the observed
data
• If the p-value is low, then that suggests that your data does not
support H0
• If the p-value is not low, then that suggests that your data does
support H0
• Rules:
• 1. If the p-value &lt; a, then we MUST REJECT H0
• 2. If the p-value &gt; a, then we CANNOT REJECT H0
• Note: The p-value is the smallest value of a for which we can
reject H0
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-24
L08
• For a greater than alternative, the p-value is the
probability of observing a value of the test statistic
greater than or equal to z when H0 is true
• Use p-value as an alternative to testing a greater
than alternative with a z test statistic (rejection
point)
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-25
L08
• Testing H0: m  50 vs.
Ha: m &gt; 50 using
rejection points in (a)
and p-value in (b)
• Have to modify steps
4, 5, and 6 of the
previous procedure to
use p-values
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-26
L09
4. Collect the sample data and compute the
value of the test statistic
• In the cable subscriptions example, the value of
the test statistic was calculated to be
z = 2.20
5. Calculate the p-value by using the test statistic
value
• The area under the standard normal curve in the
right-hand tail to the right of the test statistic z =
2.20 is the p-value
• The area is 0.5 – 0.4861 = 0.0139
• The p-value is 0.0139
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-27
L09
L10
5. Continued
• If H0 is true, the probability is 0.0139 of obtaining a
sample whose mean is \$50.575 or higher
• This is a very low probability and it appears as though
H0 is not true and should therefore be rejected
6. Rejection Rule: We will reject H0 if the p-value is less
than a
• In the cable case, a was set to 0.05
• The calculated p-value of 0.0139 is &lt; a = 0.05
• This implies that the test statistic z = 2.20 is &gt; the
rejection point z0.05 = 1.645
• Therefore reject H0 at the a = 0.05 significance level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-28
The steps are as follows:
1. State the null and alternative hypotheses
2. Specify the significance level a
3. Select the test statistic
4. Determine the rejection rule for deciding whether
or not to reject H0
5. Collect the sample data and calculate the value of
the test statistic
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
7. Interpret the statistical results in managerial terms
and assess their practical importance
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-29
L10
1. State the null and alternative hypotheses
• In the payment time case
• H0: m ≥ 19.5 vs. Ha: m &lt; 19.5, where m is the
mean bill payment time (in days)
2. Specify the significance level a
• In the payment time case, set a = 0.01
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-30
L10
3. Select the test statistic
• In the payment time case, use the test statistic
z
x  19.5
sx
x  19.5

s n
• A negative value of this this test statistic results
from a sample mean that is less than 19.5 days
•
Which provides evidence against H0 in favor of Ha
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-31
L10
4. Determine the rejection rule for deciding whether or
not to reject H0
• To decide how much less than 0 the test statistic must
be to reject H0 by setting the probability of a Type I
error to a, do the following:
• The probability a is the area in the left-hand tail of
the standard normal curve
• Use normal table to find the rejection point –za
• –za is the negative of za
• –za is the point on the horizontal axis under the standard
normal curve that gives a left-hand tail area equal to a
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-32
L10
4. Continued
• Because a = 0.01 in the payment time case, the
rejection point is –za = –z0.01 = –2.33
• Reject H0 in favor of Ha if the test statistic z is
calculated to be less than the rejection point ( –zα)
• This is the rejection rule
• In the payment time case, the rejection rule is to
reject H0 if the calculated test statistic is less than
–2.33
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-33
L10
5. Collect the sample data and calculate the value
of the test statistic
• In the payment time case, assume that s is known
and s = 4.2 days
• For a sample of n = 65, x-bar = 18.1077 days:
x  19.5 18.1077  19.5
z

 2.67
s n
4.2 65
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-34
L10
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
• Compare the value of the test statistic to the
rejection point according to the rejection rule
• In the payment time case, z = –2.67 is less than z0.01 = –2.33
• Therefore reject H0: m ≥ 19.5 in favor of
Ha: m &lt; 19.5 at the 0.01 significance level
• Have rejected H0 by using a test that allows only a 1% chance of
wrongly rejecting H0
• This result is “statistically significant” at the 0.01 level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-35
L10
7. Interpret the statistical results in managerial
terms and assess their practical importance
• We can conclude that the mean bill payment time
of the new billing system appears to be less than
19.5 days
• This is a substantial improvement over the old
system by more than 50%
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-36
L09
• Testing H0: m ≥ 19.5 vs.
Ha: m &lt; 19.5 using
rejection points in (a)
and p-value in (b)
• Have to modify steps 4,
5, and 6 of the previous
procedure to use pvalues
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-37
L09
4. Collect the sample data and compute the value of
the test statistic
• In the payment time case, the value of the test
statistic was calculated to be z = –2.67
5. Calculate the p-value using the test statistic value
• In the payment time case, the area under the
standard normal curve in the left-hand tail to the
left of the test statistic z = –2.67
• The area is 0.5 – 0.4962 = 0.0038
• The p-value is 0.0038
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-38
L10
5. Continued
• If H0 is true, then the probability is 0.0038 of obtaining
a sample whose mean is as low as 18.1077 days or
lower
• This is a very low probability and it appears as though
H0 is not true and should therefore be rejected
6. Reject H0 if the p-value is less than a
• In the payment time case, a was 0.01
• The calculated p-value of 0.0038 is &lt; a = 0.01
• This implies that the test statistic z = –2.67 is less than the
rejection point –z0.01 = –2.33
• Therefore, reject H0 at the a = 0.01 significance level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-39
• If s is known we can use a z test to test H0
• We can reject H0: m = m0 at the a level of
significance (probability of Type I error equal to a)
if and only if the appropriate rejection point rule
holds
Test Statistic
z
x  m0
s n
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
Alternative
Reject H0 if:
Ha : m &gt; m0
z &gt; -za
Ha : m &lt; m0
z &lt; -za
8-40
• If the population variance is known, we can reject
H0: m = m0 at the a level of significance (probability
of Type I error equal to a) if and only if the
corresponding p-value is less than the specified a
Test Statistic
z
x  m0
s n
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
Alternative
Reject H0 if:
Ha : m &gt; m0
p value &lt; a
Ha : m &lt; m0
p value &lt; a
8-41
• Testing a “not equal to” alternative hypothesis
• The sampling distribution of all possible sample
means is modeled by a normal curve
• Require that the true value of the population
standard deviation s is known
• What’s new and different here?
• The area corresponding to the significance level a is
evenly split into both the right- and left-hand tails
of the standard normal curve
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-42
The steps are as follows:
1. State the null and alternative hypotheses
2. Specify the significance level a
3. Select the test statistic
4. Determine the rejection rule for deciding whether
or not to reject H0
5. Collect the sample data and calculate the value of
the test statistic
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
7. Interpret the statistical results in managerial terms
and assess their practical importance
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-43
L06
1. State the null and alternative hypotheses
• In the camshaft case, H0: m = 4.5 vs.
Ha: m ≠ 4.5, where m is the mean hardness depth of
the camshaft (in mm)
2. Specify the significance level a
• In the camshaft case, set a = 0.05
3. Select the test statistic
• In the camshaft case, use the test statistic:
z
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
x  4.5
sx
x  4.5

s n
8-44
L06
3. Continued
•
A positive value of this test statistic results from a
sample mean that is greater than 4.5 mm
• Which provides evidence against H0 and for Ha
•
A negative value of this this test statistic results from a
sample mean that is less than 4.5 mm
• Which also provides evidence against H0 and for Ha
•
A very small value close to 0 (either very slightly
positive or very slightly negative) of this this test
statistic results from a sample mean that is nearly 4.5
mm
• Which provides evidence in favor of H0 and against Ha
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-45
L06
4. Determine the rejection rule for deciding
whether or not to reject H0
• To decide how different the test statistic must be
from zero (positive or negative) to reject H0 in favor
of Ha by setting the probability of a Type I error to
a, do the following:
• Divide the value of a in half to get a/2. This tail
probability will be distributed in each tail of the
standard normal curve (a two-tailed test), with
a/2 in the left tail and a/2 in the right tail
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-46
L06
4. Continued
• Use the normal table to find the rejection points
za/2 and –za/2
• za/2 is the point on the horizontal axis under the standard
normal curve that gives a right-hand tail area equal to a/2
• –za/2 is the point on the horizontal axis under the standard
normal curve that gives a left-hand tail area equal to a/2
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-47
L04
4. Continued
• Because a = 0.05 in the camshaft case, a/2=0.025
• The area under the standard normal to the right of the
mean is 0.5 – 0.025 = 0.475
• From Table A.3, the area is 0.475 for z = 1.96
• The rejection points are
za = z0.025 = 1.96 OR –za = – z0.025 = – 1.96
• Reject H0 in favour of Ha if the test statistic z satisfies
either:
1. z greater than the rejection point za/2 or
2. –z less than the rejection point –za/2
&raquo; This is the rejection rule
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8-48
L04
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-49
L04
5. Collect the sample data and calculate the value
of the test statistic
• In the camshaft case, assume that s is known and
s = 0.47 mm
• For a sample of n = 40, x-bar = 4.26 mm
• Then
z
x  4.5
s n
Copyright &copy; 2011 McGraw-Hill Ryerson Limited

4.26  4.5
0.47
 3.02
35
8-50
L06
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
• Compare the value of the test statistic to the
rejection point according to the rejection rule
• In the camshaft case, z = –3.02 is less than
– z0.025 = –1.96
• Therefore we must reject H0: m = 4.5 in favor of
Ha: m ≠ 4.5 at the 0.05 significance level
• We have rejected H0 using a test that allows for only a 5%
chance of wrongly rejecting H0
• This result is said to be “statistically significant” at the 5% level
of significance
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-51
L06
7. Interpret the statistical results in managerial
terms and assess their practical importance
• We can conclude that it appears as though the
mean hardness depth of the camshaft is not 4.5
mm
• Using x = 4.26 and s = 0.47, we estimate that
99.73 percent of all individual camshaft hardness depths are in the
interval [ x  3s ]  4.26  30.47)  2.85,5.67
• This estimated tolerance interval says that, because of a somewhat
small mean of 4.26, we estimate that we are producing some
hardness depths that are below the lower specification limit of 3
mm
• Changes must be made to the manufacturing process
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-52
L04
• Testing H0: m = 4.5 versus Ha: m ≠ 4.5 for a = 0.05
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-53
L09
4. Collect the sample data and compute the value of
the test statistic
• In the camshaft case, the value of the test statistic
was calculated to be z = –3.02
5. Calculate the one-sided p-value using the test
statistic value
• In the camshaft case, the area under the standard
normal curve in the left-hand tail to the left of the
test statistic value z = –3.02
• The area is 0.5 – 0.4987 = 0.0013
• The one-sided p-value is 0.0013
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8-54
L09
5. Continued
• We are conducting a two-sided test though, and therefore, need a
two-sided p-value. All we need to do is double the one-sided pvalue
• Our two-sided p-value would be 2(0.0013) = 0.0026
• This is a very low probability and it appears as though H0 is not
true and should therefore be rejected
6. Reject H0 if the two-sided p-value is less than a
• In the camshaft case, a was 0.05
• The calculated two-sided p-value of 0.0026 is &lt; a = 0.05
• Therefore reject H0 at the a = 0.05 significance level
• We would also reject H0 at the 1% significance level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-55
L09
• Using a z statistic to test H0 when the population
variance is known, we can reject H0: m = m0 at the
a level of significance (probability of Type I error
equal to a) if and only if:
Test Statistic
Alternative
Reject H0 if:
z &gt; za/2
x  m0
z
s n
Ha : m ≠ m0
(1)
or
2-sided p-value &lt;
a
z &lt; –za/2
(2)
or
2-sided p-value &lt;
a
1.
2.
a/2 is the area under the standard normal curve to the right of za/2
a/2 is the area under the standard normal curve to the left of –za/2
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-56
L09
• Calculate the test statistic and the corresponding p-value
• You can rate the strength of the conclusion, in general, about H0
according to these guidelines:
• With a p-value &lt; 0.10, there is some evidence to reject H0
• With a p-value &lt; 0.05, there is strong evidence to reject H0
• With a p-value &lt; 0.01, there is very strong evidence to reject H0
• With a p-value &lt; 0.001, there is extremely strong evidence to reject
H0
• So, the smaller the p-value, the stronger the evidence is against Ho
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
8-57
• The null hypothesis H0 can be rejected in favor of the
alternative Ha at some level of significance, a, if and only
if the 100(1 – a)% confidence interval for m does not
contain m0
• where m0 is the claimed value for the population mean
• In other words, if the confidence interval does not
contain the claimed value m0, then the null hypothesis
can be rejected
• Conversely, if the confidence interval does contain the
claimed value m0, then the null hypothesis cannot be
rejected
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•In the camshaft case, the 95% confidence interval for m
is 
s  
0.47 
 x  za 2
  4.26  1.96
  4.10 , 4.42 

n

35 
•which does not contain target value of 4.5 mm
•Therefore we must reject the null hypothesis that m is
4.5 mm at the 0.05 significance level
•Here, the confidence level 1 – a = 0.95 and the significance level a = 0.05
•These sum to 1, so here see that the confidence level and significance levels
are complementary
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• The a we use here is the same a that we used in
Chapter 7
• The confidence level is 1 – a
• The significance level is a
• (1 – a ) + a = 1, so the confidence level and
significance levels are complementary
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• Suppose the population being sampled is
normally distributed
• The population standard deviation s is
unknown, as is the usual situation
• If the population standard deviation s is
unknown, then it will have to estimated from a
sample standard deviation s
• Under these two conditions, have to use the
t distribution to test hypotheses
• The t distribution from Chapter 7
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• Define a new random variable t:
t
x m
s
n
• with the definition of symbols as before
• The sampling distribution of this random variable
is a t distribution with n – 1 degrees of freedom
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• Let x-bar be the mean of a sample of size n with
standard deviation s
• Also, m0 is the claimed value of the population mean
• Define a new test statistic
t
x  m0
s
n
• If the population being sampled is normal, and
• If s is used to estimate s, then …
• The sampling distribution of the t statistic is a t
distribution with n – 1 degrees of freedom
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• Same rejection rules apply here that applied with
our Z-tests
• Reject H0: m = m0 in favor of a particular alternative
hypothesis Ha at the a level of significance if and
only if the test statistic lies in the appropriate
rejection region or, equivalently, if the
corresponding p-value is less than a
• We have the following rules …
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L09
Alternative Reject H0 if: p-value (reject Ho if p-value &lt; a)
Ha: m &gt; m0
t &gt; ta
Area under t distribution to right of t
Ha: m &lt; m0
t &lt; –ta
Area under t distribution to left of –t
Ha: m  m0
|t| &gt; t a/2 *
Twice area under t distribution to
right of |t|
• ta, ta/2, and p-values are based on n – 1 degrees of freedom (for a
sample of size n)
• * either t &gt; ta/2 or t &lt; –ta/2
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8-65
• In 2001, the mean credit card interest rate is as it was
at the rate of 18.8%
• A financial officer wishes to study whether the increased
competition in the credit card business has reduced
interest rates
• This is the null hypothesis H0: m = 18.8
• The alternative to be tested is that the current mean
interest rate isn’t as it was back in 2001, but in fact has
decreased
• This is the alternative hypothesis Ha: m &lt; 18.8
• Test at the a = 0.05 level of significance
• If H0 can be rejected at the 5% level, then conclude that
it appears as though the mean rate now is less than
18.8% rate charged in 2001
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L04
• We randomly select n = 15 credit cards
• We know:
n = 15
x = 16.828
s = 1.538
a = 0.05
• Then df = n – 1 = 14
• We will assume the population of the rates of all
cards is approximately normal
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L04
• The rejection rule is:
Reject H0: m = 18.8 against Ha: m &lt; 18.8 if t &lt; ta
• For a = 0.05 and with df = 14, ta = t0.05 =1.761
• Calculate the value of the t statistic:
t
x  18.8
s n

16.827  18.8
1.538 15
 4.97
• Because t = 4.97 &lt; t0.05 = 1.761, reject H0
• Therefore, conclude at 5% significance level that it
appears as though the mean interest rate is lower than
it was in 2001
• In fact it was observed to be 18.8 – 16.827 = 1.973%
lower
• Our observed result of 16.827 is said to be “statistically
significant” at a = 5%
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
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L09
L10
• The p-value here is the left-hand tail area under the t curve with df
= 14 to the left of t = 4.97
• Our observed t value is off the t table, so using the table, we would
say that our p-value &lt; 0.0005.
• A statistical software package can easily calculate the exact pvalue. It turns out that the p-value is actually 0.000103, which is
less than 0.0005.
• This says that if the claimed mean of 18.8 % is true, then there is
only about a 1/10,000 chance of randomly selecting a sample of 15
credit cards whose mean rate would be as low as or lower than
16.827%
• This p-value is &lt; 0.05, 0.01, 0.001
• We would reject H0 at the 0.05, 0.01, 0.001 significance levels
• Our result is highly significant
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L09
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• If the sample size n is large, we can reject H0: p = p0
at the a level of significance (probability of Type I
error equal to a) if and only if the appropriate
rejection point condition holds or, equivalently, if
the corresponding p-value is less than a
• We have the following rules …
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Alternative
Reject H0 if:
p-value
Ha: p &gt; p0
z &gt; zα
Area under standard normal to the right of z
Ha: p &lt; p0
z &lt; -zα
Area under standard normal to the left of –z
Ha: p ≠ p0
|z| &gt; zα/2 *
Twice the area under standard normal to the
right of |z|
• The test statistic is: z 
p̂  p0
p0 1 p0 )
n
* either z &gt; za/2 or z &lt; –za/2
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
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L06
• Virol has been shown to provide relief for 70% of
all patients suffering from upper respiratory
infections. A drug company believes that its
competing drug, Phantol, is more effective than
Virol
• Let’s test Ho at a = 5% using rejection points and pvalues.
• Test H0: p = 0.70 versus Ha: p &gt; 0.70
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L06
• A random sample of 300 patients was taken and
231 indicated that the drug had effectively treated
their upper respiratory infections
231
 0.77
• So, we have n = 300 and p 
300
^
• In order to test Ho, we need a test statistic:
z
ˆp  p 0
p 0 1  p 0 )
n
Copyright &copy; 2011 McGraw-Hill Ryerson Limited

0.77 - 0.70
0.70 1  0.70)
300
 2.65
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L06
• Reject H0: p = 0.70 if the value of Z is greater than
zA = z0.05 = 1.645
• Note that using this procedure is valid because np0 = 300(0.70) =
210 and n(1 - p0) = 300(1 - 0.70) = 90 are both at least 5
• Comparing values:
• z = 2.65 &gt; z0.05 = 1.645
&gt; z0.01 = 2.33
&lt; z0.001 = 3.09
• So reject H0 at the 0.05 and 0.01 significance
levels but do not reject at the 0.001 level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
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L09
• For z = 2.65, the p-value is:
p-value = P(z &gt; 2.65) = (0.5 – 0.4960) = 0.004
(table A.3)
• But p = 004 &lt; a = 0.05
&lt; a = 0.01
&gt; a = 0.001
• So (again) reject H0 at the 0.05 and 0.01
significance levels but do not reject at the 0.001
level
Copyright &copy; 2011 McGraw-Hill Ryerson Limited
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• Hypothesis tests are used to test the strength of a
statement for a population
• The null hypothesis H0 is the statement being tested, it is
the status quo, it is not rejected for the alternative
hypothesis Ha unless there is convincing statistical evidence
otherwise
• There are three common ways to test a hypothesis 1)
rejection region 2) p-value 3) confidence intervals
• P-values can give more information as it is the value of α at
which the decision will change
• Types of parameters which we test in this chapter are
means, proportions
• Type I errors occur when a true null hypothesis is rejected
and type II errors occur when a false null hypothesis is not
rejected
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